Eigenvalues for an Invertible Matrix

[Doesy]

Homework Statement

A is an invertible matrix, x is an eigenvector for A with an eiganvalue \lambda \neq0 Show that x is an eigenvector for A^-1 with eigenvalue \lambda^-1

Homework Equations

Ax=\lambdax
(A - I)x

The Attempt at a Solution

I know that I need to find x and then apply to the inverses of my Matrix and eigenvalue, but how do I know what matrix to use for A? Do I use the inverse matrix as it is an invertible matrix? Can I use any invertible matrix to prove this?

Thanks in advance.

 

[Cyosis]

Let \mu be the eigenvalue of A^{-1}. Now multiply the eigenvalue equation with A^{-1} and find \mu.

 

[Doesy]

Do you mean I should have

A^{2}x = A\lambdax

and

A^{-2}x = A^{-1}\mux

?

How can I use this to show my Answer? Or do I substitute this second equation into the first?

 

[Cyosis]

No with multiplying the eigenvalue equation I meant A^{-1}(Ax)=A^{-1}\lambda x. On a side note, use brackets around your entire equation not just lambda or mu.

 

[Doesy]

So I can re-arrange as
\frac{A^{-1}(A\chi)}{\lambda} = \mu\chi
Is that right?

 

[Cyosis]

Sure, because the problem says \lambda \neq 0.What is A^{-1}A?

 

[Doesy]

Is A^-1*A the Inverse Matrix I?

 

[Cyosis]

Yes it's the identity matrix, which has itself as an inverse. What is Ix?

 

[Doesy]

Inverse multiplied by the eigenvector will give us the original matrix?

 

[Cyosis]

Are you calling I the inverse? It's the Identity matrix. If you let the Identity matrix work on a vector it yields that vector without making any changes to it. We're moving in the wrong direction now so let's start at the eigenvalue equation again. A^{-1}(Ax)=A^{-1}\lambda x \Rightarrow Ix=\lambda \mu x. Now solve for \mu.

 

[Doesy]

Oops, my bad, I meant Identity.

Ix gives us x again.

So:

\frac{I\chi}{\lambda\chi} = \mu

Is that right? and Ix = x?

 

[Cyosis]

Well x are vectors how do you define a vector divided by another vector? While doing it in this case and letting the xs cancel you will get the correct answer, I would suggest you don't do this on your exam.

This is what you should do, Ix=\lambda \mu x \Rightarrow x=\lambda \mu x \Rightarrow 1*x=\lambda \mu x. So what is \mu?

 

[Doesy]

\mu = \frac{1}{\lambda}

Correct?

 

[Cyosis]

Yes, perhaps it's nice if you write the full proof down now so we can see if you don't do any operations that you shouldn't really be doing.

 

[Doesy]

Alright Cool, Here Goes!

A\chi = \lambda\chi

A^{-1}\chi = \lambda^{-1}\chi

Let \lambda^{-1} = \mu

<br /> A^{-1}(Ax)=A^{-1}\lambda x<br />

We can re-write this as

<br /> \frac{A^{-1}(A\chi)}{\lambda} = \mu\chi<br />

<br /> A^{-1}A = I<br />

Here I is the Identity Matrix

Ix = x

As The Identity matrix multiplied by a vector value does not change the position of the vector. So we now have:

<br /> \frac{I\chi}{\lambda\chi} = \mu<br />

This equates to:

\frac{1}{\lambda\chi}

This shows that the vector \chi is an eigen vector for the Inverse Matrix of A as it's Eigen Value is also the Inverse of \chi

 

[Cyosis]

Alright Cool, Here Goes!

A\chi = \lambda\chi

A^{-1}\chi = \lambda^{-1}\chi

Let \lambda^{-1} = \mu

You need to proof that \mu=\lambda^{-1}, not start with it.

What we are given is the eigenvalue equation Ax=\lambda x with \lambda being an eigenvalue of A. We want to proof that x is an eigenvector of A^{-1} with eigenvalue \lambda^{-1}.
So let's multiply the eigenvalue equation by A^{-1}.
This yields
A^{-1}(Ax)=A^{-1} \lambda x \Rightarrow (A^{-1}A)x=\lambda A^{-1} x \Rightarrow Ix=\lambda A^{-1}x \Rightarrow 1*x=\lambda A^{-1} x \Rightarrow \lambda^{-1}x=A^{-1}x.
A matrix working on a vector that yields the same vector multiplied by a scalar is an eigenvalue equation. So x must be an eigenvector of A^{-1} and \lambda^{-1} must be an eigenvalue of A^{-1}. There is no dividing of any vectors happening here! If you want to divide two vectors you must first define how division with vectors works.

 

[Doesy]

Ohhhh!

Thanks heaps man, is there anyway I can +rep you or something?